Quantity of gasoline needed by a car to run 800 miles = 30 gallons
Quantity of gasoline needed by a car to run 1 mile =
= 30 ÷ 800
= 0.0375 gallons
So , to run 1 mile a car would need = 0.0375 gallons of oil
To run 700 miles the quantity of gasoline needed =
= 700 × 0.0375
= 26.25 gallons of gasoline
Therefore , a car will use 26.25 gallons of gasoline on a trip of 700 miles .
Given equation : y=5/2x.
Solution: In order to graph direct variation equation, we need to make a table of different values of x and y.
We need to plug some values of x's in equation to get values of y's.
Because, we have fraction 5/2, and 2 in denominator. So it would be better to take multiples of 2, so that we don't get values in decimals.
Let us take x=0 first.
Plugging x=0 in given equation, we get
y=5/2x = 5/2(0) = 0.
Let us take x=2 first.
Plugging x=2 in given equation, we get
y=5/2 *(2) = 5
Let us take x=2 first.
Plugging x=4 in given equation, we get
y=5/2 *(4) = 5 *2 = 10.
We got three coordinates for the graph (0,0) , (2,5) and (4, 10).
Let us plot those coordinates on the graph and join the points to draw a straight line.
Given:
Angle A = 18.6°
Angle B = 93°
Length of side AB = 646 meters
To find:
the distance across the river, distance between BC
Steps:
Since we know the measure of 2 angles of a triangle we can find the measure of the third angle.
18.6° + 93° + ∠C = 180°
111.6° + ∠C = 180°
∠C = 180° - 111.6°
∠C = 68.4°
Therefore the measure of angle C is 68.4°.
now we can use the law of Sines,


![BC[sin(68.4)] = 646 [sin(18.6)]](https://tex.z-dn.net/?f=BC%5Bsin%2868.4%29%5D%20%3D%20646%20%5Bsin%2818.6%29%5D)



meters
Therefore, the distance across the river is 222 meters.
Happy to help :)
If anyone need more help, feel free to ask
X(-x^4)(-x^4)=
The answer is x^9